Mixing
The convolution of functions in the Schwartz Space lies in the Schwartz Space
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I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:
In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:
$$|x|^l |g(x-y)| leq A_l(1 + |y|)^l forall l geq 0$$ Remember, $S(mathbbR)$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^(k)(x)| < infty forall k,l$
The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.
Thanks
inequality fourier-analysis convolution schwartz-space
asked Jul 17 at 11:22
Ale.B
628
add a comment |Â
up vote
4
down vote
favorite
I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:
In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:
$$|x|^l |g(x-y)| leq A_l(1 + |y|)^l forall l geq 0$$ Remember, $S(mathbbR)$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^(k)(x)| < infty forall k,l$
The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.
Thanks
inequality fourier-analysis convolution schwartz-space
asked Jul 17 at 11:22
Ale.B
628
In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:
In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:
$$|x|^l |g(x-y)| leq A_l(1 + |y|)^l forall l geq 0$$ Remember, $S(mathbbR)$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^(k)(x)| < infty forall k,l$
The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.
Thanks
inequality fourier-analysis convolution schwartz-space
asked Jul 17 at 11:22
Ale.B
628
I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:
In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:
$$|x|^l |g(x-y)| leq A_l(1 + |y|)^l forall l geq 0$$ Remember, $S(mathbbR)$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^(k)(x)| < infty forall k,l$
The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.
Thanks
inequality fourier-analysis convolution schwartz-space
asked Jul 17 at 11:22
Ale.B
628
asked Jul 17 at 11:22
Ale.B
628
asked Jul 17 at 11:22
Ale.B
628
asked Jul 17 at 11:22
Ale.B
628
628
In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18
add a comment |Â
In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18
In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Recall the following trivial inequality;
$$lvert xrvert ^l leq left(, lvert x-y rvert , + lvert y rvert , right)^lleq (2max(, lvert x-yrvert, , lvert y rvert , ), )^l leq, 2^l , lvert x-yrvert^l + 2^l, lvert y rvert ^l qquad forall lgeq 0$$
Set $$C_l := sup_xinmathbbR , lvert xlvert ^l ,lvert g(x) , rvert <infty qquad forall lgeq 0$$
Now using the above estimate, we get $$lvert x rvert^l , lvert g(x-y) rvert leq 2^l , lvert x-yrvert^l , rvert g(x-y)rvert + 2^l, lvert y rvert^l , lvert g(x-y) rvert leq \ 2^l, C_l+ 2^l , C_0 , lvert y rvert^l leq underbrace2^l(C_l+C_0)_=A_l, (1+lvert yrvert)^l$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ lvert x rvert^l , lvert (gast f)(x) rvert leq int _mathbbR lvert x rvert^l , lvert g(x-y) rvert , lvert f(y) rvert , dy leq A_l int_mathbbR (1+lvert y rvert )^l+2 , lvert f(y) rvert fracdy(1+lvert y rvert)^2 leq \ A_l , C_l+2 ,int_mathbbR fracdy(1+lvert y rvert)^2< infty $$
Since the bound is independent of $xin mathbbR$, the claim now follows.
answered Jul 18 at 15:49
TheOscillator
2,0441716
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall the following trivial inequality;
$$lvert xrvert ^l leq left(, lvert x-y rvert , + lvert y rvert , right)^lleq (2max(, lvert x-yrvert, , lvert y rvert , ), )^l leq, 2^l , lvert x-yrvert^l + 2^l, lvert y rvert ^l qquad forall lgeq 0$$
Set $$C_l := sup_xinmathbbR , lvert xlvert ^l ,lvert g(x) , rvert <infty qquad forall lgeq 0$$
Now using the above estimate, we get $$lvert x rvert^l , lvert g(x-y) rvert leq 2^l , lvert x-yrvert^l , rvert g(x-y)rvert + 2^l, lvert y rvert^l , lvert g(x-y) rvert leq \ 2^l, C_l+ 2^l , C_0 , lvert y rvert^l leq underbrace2^l(C_l+C_0)_=A_l, (1+lvert yrvert)^l$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ lvert x rvert^l , lvert (gast f)(x) rvert leq int _mathbbR lvert x rvert^l , lvert g(x-y) rvert , lvert f(y) rvert , dy leq A_l int_mathbbR (1+lvert y rvert )^l+2 , lvert f(y) rvert fracdy(1+lvert y rvert)^2 leq \ A_l , C_l+2 ,int_mathbbR fracdy(1+lvert y rvert)^2< infty $$
Since the bound is independent of $xin mathbbR$, the claim now follows.
answered Jul 18 at 15:49
TheOscillator
2,0441716
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
add a comment |Â
up vote
2
down vote
accepted
Recall the following trivial inequality;
$$lvert xrvert ^l leq left(, lvert x-y rvert , + lvert y rvert , right)^lleq (2max(, lvert x-yrvert, , lvert y rvert , ), )^l leq, 2^l , lvert x-yrvert^l + 2^l, lvert y rvert ^l qquad forall lgeq 0$$
Set $$C_l := sup_xinmathbbR , lvert xlvert ^l ,lvert g(x) , rvert <infty qquad forall lgeq 0$$
Now using the above estimate, we get $$lvert x rvert^l , lvert g(x-y) rvert leq 2^l , lvert x-yrvert^l , rvert g(x-y)rvert + 2^l, lvert y rvert^l , lvert g(x-y) rvert leq \ 2^l, C_l+ 2^l , C_0 , lvert y rvert^l leq underbrace2^l(C_l+C_0)_=A_l, (1+lvert yrvert)^l$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ lvert x rvert^l , lvert (gast f)(x) rvert leq int _mathbbR lvert x rvert^l , lvert g(x-y) rvert , lvert f(y) rvert , dy leq A_l int_mathbbR (1+lvert y rvert )^l+2 , lvert f(y) rvert fracdy(1+lvert y rvert)^2 leq \ A_l , C_l+2 ,int_mathbbR fracdy(1+lvert y rvert)^2< infty $$
Since the bound is independent of $xin mathbbR$, the claim now follows.
answered Jul 18 at 15:49
TheOscillator
2,0441716
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall the following trivial inequality;
$$lvert xrvert ^l leq left(, lvert x-y rvert , + lvert y rvert , right)^lleq (2max(, lvert x-yrvert, , lvert y rvert , ), )^l leq, 2^l , lvert x-yrvert^l + 2^l, lvert y rvert ^l qquad forall lgeq 0$$
Set $$C_l := sup_xinmathbbR , lvert xlvert ^l ,lvert g(x) , rvert <infty qquad forall lgeq 0$$
Now using the above estimate, we get $$lvert x rvert^l , lvert g(x-y) rvert leq 2^l , lvert x-yrvert^l , rvert g(x-y)rvert + 2^l, lvert y rvert^l , lvert g(x-y) rvert leq \ 2^l, C_l+ 2^l , C_0 , lvert y rvert^l leq underbrace2^l(C_l+C_0)_=A_l, (1+lvert yrvert)^l$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ lvert x rvert^l , lvert (gast f)(x) rvert leq int _mathbbR lvert x rvert^l , lvert g(x-y) rvert , lvert f(y) rvert , dy leq A_l int_mathbbR (1+lvert y rvert )^l+2 , lvert f(y) rvert fracdy(1+lvert y rvert)^2 leq \ A_l , C_l+2 ,int_mathbbR fracdy(1+lvert y rvert)^2< infty $$
Since the bound is independent of $xin mathbbR$, the claim now follows.
answered Jul 18 at 15:49
TheOscillator
2,0441716
Recall the following trivial inequality;
$$lvert xrvert ^l leq left(, lvert x-y rvert , + lvert y rvert , right)^lleq (2max(, lvert x-yrvert, , lvert y rvert , ), )^l leq, 2^l , lvert x-yrvert^l + 2^l, lvert y rvert ^l qquad forall lgeq 0$$
Set $$C_l := sup_xinmathbbR , lvert xlvert ^l ,lvert g(x) , rvert <infty qquad forall lgeq 0$$
Now using the above estimate, we get $$lvert x rvert^l , lvert g(x-y) rvert leq 2^l , lvert x-yrvert^l , rvert g(x-y)rvert + 2^l, lvert y rvert^l , lvert g(x-y) rvert leq \ 2^l, C_l+ 2^l , C_0 , lvert y rvert^l leq underbrace2^l(C_l+C_0)_=A_l, (1+lvert yrvert)^l$$
This establishes the desired inequality. To finish off the proof entirely we may write $$ lvert x rvert^l , lvert (gast f)(x) rvert leq int _mathbbR lvert x rvert^l , lvert g(x-y) rvert , lvert f(y) rvert , dy leq A_l int_mathbbR (1+lvert y rvert )^l+2 , lvert f(y) rvert fracdy(1+lvert y rvert)^2 leq \ A_l , C_l+2 ,int_mathbbR fracdy(1+lvert y rvert)^2< infty $$
Since the bound is independent of $xin mathbbR$, the claim now follows.
answered Jul 18 at 15:49
TheOscillator
2,0441716
edited Jul 18 at 21:43
answered Jul 18 at 15:49
TheOscillator
2,0441716
answered Jul 18 at 15:49
TheOscillator
2,0441716
answered Jul 18 at 15:49
TheOscillator
2,0441716
2,0441716
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
add a comment |Â
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
I am not too sure about some of the inequalities.
– Ale.B
Jul 18 at 20:29
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l leq 2^l |x-y|^l + 2^l |y|^l$.
– Ale.B
Jul 18 at 20:35
1
1
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
Ooooh! I get it!!!
– Ale.B
Jul 18 at 21:10
add a comment |Â
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In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz...
– paul garrett
Jul 18 at 22:25
Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R).
– Ale.B
Jul 19 at 3:14
well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven...
– paul garrett
Jul 19 at 12:18